Chapter 3: Problem 41
Calculate the linearization \(L(x)=f(c)+\) \(f^{\prime}(c), \cdot(x-c)\) for the given function \(f\) at the given value \(c\) $$ f(x)=e(x-1) / x, c=1 $$
Short Answer
Expert verified
The linearization of the function at \( c = 1 \) is \( L(x) = 1 \).
Step by step solution
01
Identify the Function and the Point of Interest
The function given is \( f(x) = \frac{e^{x-1}}{x} \), and we need to find the linearization at the point \( c = 1 \).
02
Evaluate the Function at \( c \)
Substitute \( c = 1 \) into the function to find \( f(1) \):\[f(1) = \frac{e^{1-1}}{1} = \frac{e^0}{1} = 1.\]
03
Find the Derivative of the Function
Use the quotient rule to differentiate \( f(x) = \frac{e^{x-1}}{x} \):The quotient rule is \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).Let \( u = e^{x-1} \) with \( u' = e^{x-1} \),and \( v = x \) with \( v' = 1 \).\[f'(x) = \frac{(e^{x-1})(x) - (e^{x-1})(1)}{x^2} = \frac{xe^{x-1} - e^{x-1}}{x^2} = \frac{e^{x-1}(x-1)}{x^2}.\]
04
Evaluate the Derivative at \( c \)
Substitute \( c = 1 \) into \( f'(x) \):\[f'(1) = \frac{e^{1-1}(1-1)}{1^2} = \frac{e^0(0)}{1} = 0.\]
05
Formulate the Linearization \( L(x) \)
The linearization formula is \( L(x) = f(c) + f'(c)(x - c) \). Substitute the values found:\[L(x) = 1 + 0(x - 1) = 1.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
The quotient rule is a fundamental tool in calculus used to find the derivative of a quotient of two functions. Suppose you have a function expressed as a ratio, say \( f(x) = \frac{u}{v} \), where \( u \) and \( v \) are functions of \( x \). To differentiate \( f \), we use the quotient rule:
- The formula for the quotient rule is: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).
- Here, \( u' \) is the derivative of \( u \), and \( v' \) is the derivative of \( v \).
- The numerator involves the difference between two products: \( u'v \) and \( uv' \).
- The denominator is simply \( v^2 \).
Differentiation
Differentiation is a process that estimates the rate at which a function changes. It's a crucial concept in calculus, particularly in determining slopes of tangent lines to curves. When you differentiate a function, you are essentially finding its derivative.
Purpose of Differentiation
- Calculates the rate of change or the slope at any point of the function.
- Essential in physics for motion and dynamics, and in economics for finding trends and optimizing functions.
- Helps in solving real-world problems where change is involved.
Function Evaluation
Function evaluation involves substituting specific values into a function to determine its output. It's one of the fundamental methods for understanding how a function behaves at a certain point.
Steps in Function Evaluation
- Identify the point at which you need to evaluate the function, for instance, \( c = 1 \).
- Substitute this value into the function to find the result. For \( f(x) = \frac{e^{x-1}}{x} \), evaluating at \( x = 1 \) gives \( f(1) = 1 \).
- Ensure correct arithmetic: work systematically to avoid mistakes.